Page 43 - Maths Class 06

For greatest digit,
To make 5 729p divisible by 3, we will add multiples of 3 is 3, 6, 9 to p = 1
i e. . 1 + 3 = 4

1 + 6 = 7
1 + 9 10=
Since, p cannot be two digit number, the greatest value of p = 7.

Hence, 57729 is required greatest number.

Exercise 3.2

1. Check divisibility of the follow ing numbers by 2, 4, 8, 5 and 10. Put a tick (3) for divis i ble and
cross (7) for not divisible:

Number 2 4 8 5 10

(a) 990

(b) 464
(c) 572

(d) 1586

(e) 4995

(f) 94505

(g) 66660
(h) 639210

(i) 726352

2. Check divisibility of the follow ing numbers by 3, 6, 9 and 11. Put a tick (3) for divis i ble and
cross (7) for not divisible:

Number 3 6 9 11

(a) 1258
(b) 5335

(c) 71232

(d) 10824

(e) 21084

(f) 639216

(g) 901351

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